Fourier Heat Equation

$\nabla (k\nabla T) + \dot{q}_{p} = \rho c_{p} \frac{dT}{dt}$

Here, $k$ is the thermal conductivity ($W/(m·K)$), $\rho$ is the density ($kg/m^{3}$), and $c_{p}$ is the specific heat capacity ($J/(kg·K)$).

Consider the case of an infinite plate with the initial temperature $T_{0} = 300K$ immersed in a liquid that is at a different temperature $T_{f}$. In this case, transfer of heat takes place till the entire plate reaches the temperature $T_{f}$. We need to see how the temperature varies along the width of the plate, and how this temperature profile changes over time, depending on the $T_{f}$ and the property of the material, for example, the specific heat capacity $c_{p}$.

Approach to the solution (1/2)

Given that we have an infinite plate, only one dimension (along the thickness) is of interest to us. Further, assuming no external heat, we can put $\dot{q} = 0$. Taking the thermal conductivity to be constant. So, our equation becomes,

$k\frac{d^2T}{dx^2} = \rho c_{p} \frac{dT}{dt}$.

The initial condition is: $T(x,0) = T_{0}$. The boundary conditions are:

1. $\frac{dT(0,t)}{dx} = 0$
2. $-k\frac{dT(L,t)}{dx} = h[T(L,t)-T_{f}]$

For making the equation dimensionless and simplifications, we use: $Bi = \frac{hL}{k}$; $\alpha = \frac{k}{c_{p}\rho}$; $\tau = \frac{\alpha t}{L^2}$ and $X = \frac{x}{L}$

Approach to the solution (2/2)

So, after separation of variables, the one-dimensional Fourier equation becomes: $\frac{1}{F}\frac{d^2F}{dX^2} = \frac{1}{G}\frac{d^2G}{d\tau^2}$

Let this be equal to $-\lambda_{n}^{2}$. Now, we apply the boundary conditions, and solve for $\lambda_{n}$. This simplifies to $\lambda_n tan(\lambda_{n}) = Bi$ -- (1).

For a given material, changing the final Temperature $T_f$

In this case, $Bi$, and $\alpha$ remain unchanged. We have considered a material with $c_p = 390 J/(kg.K)$ kept at an initial temperature of 300 K. For simplicity, we consider the case where $\tau > 0.2$ and we consider only the first solution of equation 1.

fig = px.line_3d(df, x='Distance', y='Time', z='Temperature', animation_frame='Final temperature',range_x=[-5,5],range_z=[300,450], title = 'Temperature (K) profile with varying final temperature')
fig.show()

For a given Final Temperature, changing the material's property $c_p$

In this case, $Bi$, and $\alpha$ change. We have considered a material with a starting $c_p$ of $390 J/(kg.K)$ with initial temperature of 300K and final temperature of 400K. Again, for simplicity, we consider the case where $\tau > 0.2$ and we consider only the first solution of equation 1.

fig = px.line_3d(df, x='Distance', y='Time', z='Temperature', animation_frame='Material Property',range_x=[-5,5],range_z=[300,410], title = 'Temperature (K) profile with varying Specific Heat Capacity')
fig.show()